Question 1.213: Let R ⊆ S be a ring extension such that S is an integral dom......

Suppose R is a field. Since S is an integral extension of R, if s \in S, there exists a finitely generated R-submodule T of S such that s \in T and T is a ring. In particular, T is a finite dimensional vector space over R, hence 1, s, s^{2}, \ldots is linearly dependent:

there exists a_{0}, a_{1}, \ldots, a_{m} \in R such that \sum^{m} a_{i} s^{i}=0. Without loss of generality we assume a_{0} \neq 0. Thus, -a_{0}=s\left(a_{1}+\cdots+a_{m} s^{m-1}\right), so s is invertible, hence S is a field. Conversely, if S is a field and r \in R, then we are going to show that 1 / r belongs to R. Indeed, since 1 / r \in S and S is integral over R, a_{0}+\frac{1}{r} a_{1}+\cdots+\frac{1}{r^{m}} a_{m}+\frac{1}{r^{m+1}}=0 for some a_{i} \in R, i=0, \ldots, m. Equivalently, a_{0} r^{m}+a_{1} r^{m-1}+\cdots+a_{m}=-\frac{1}{r}, which implies that 1 / r \in R.